Problem: Solve for $r$, $ -\dfrac{4}{4r + 10} = -\dfrac{r - 8}{2r + 5} + \dfrac{1}{2r + 5} $
Solution: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4r + 10$ $2r + 5$ and $2r + 5$ The common denominator is $4r + 10$ The denominator of the first term is already $4r + 10$ , so we don't need to change it. To get $4r + 10$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ -\dfrac{r - 8}{2r + 5} \times \dfrac{2}{2} = -\dfrac{2r - 16}{4r + 10} $ To get $4r + 10$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{1}{2r + 5} \times \dfrac{2}{2} = \dfrac{2}{4r + 10} $ This give us: $ -\dfrac{4}{4r + 10} = -\dfrac{2r - 16}{4r + 10} + \dfrac{2}{4r + 10} $ If we multiply both sides of the equation by $4r + 10$ , we get: $ -4 = -2r + 16 + 2$ $ -4 = -2r + 18$ $ -22 = -2r $ $ r = 11$